I love the electrical analogy, it reminds me strongly of a relaxation oscillator and if I remember correctly it was Larry who first put forward that particular circuit as an analogy, correct me if I am wrong Larry, it does strongly suggest (with your other post on valves) that there is a very narrow band of conditions were the combustor will operate.

Viv

"Sometimes the lies you tell are less frightening than the loneliness you might feel if you stopped telling them"Brock Clarke

Many years ago, I took an introductory course in System Dynamics and the crux of the course was to convert elements of various disciplines (heat transfer, mechanics, fluids, etc.) to electrical analogs. Why?

In the early days of computation, computers were of the analog type. To solve an equation one applied voltage to array of resistors, capacitors, inductors, diodes, and such and measured out a voltage, probably on an oscilloscope, for the solution. To obtain a solution to an equation from some other field of study than electrical/electronics, one had to properly transform his/her equation to an electrical analogy.

It was one of those things that had to be learned. It is just one more 'trick' in an engineer/scientist's tool bag. Kudos to Larry if he was the first to post. I hope he took the time to solve it, because I'd like to compare results.

Anyways, I solved this problem using a program I wrote for a digital computer. I still used Kirchoff's current and voltage laws in developing the governing equations, though.

Particulars:
[*]dfr is 0.5
[*]it appears that the valve closes exactly at the end of the -tive portion of the cycle pressure (see arrow)
[*]max amplitude of theta is comparible to that found in dfr=1.0
[*]my problem with this valve is that the movement of the valve 'lags' the pressure across the orifice - in other words, at the end of the -tive portion of the cycle pressure, the valve is still open but the pressure, p12, across the orifice is rapidly approaching zero. UNSAT

Ape wrote:This is propably a rather silly question, showing my complete lack of understanding of what you're trying to relate here, but, well, here goes ; How can dfr be >1?
What did I miss?

Hi Ape,

No harm done in asking.

For the sake of illustration, let's assume my pump runs at 1800 rpm.

f = (1800 rev/min) / (60 sec/min) gives 30 rev/sec

During 1 rev the pump cycles 1 time so

f = 30 cycles/sec for the pump frequency

let fn = resonant frequency of the valve

I have defined driving frequency ratio as

dfr = w / wn = (2pi/2pi) x w / wn = f / fn

So, the dfr is dependent on the resonant frequency of the valve, fn

If fn = 10 cps, then dfr = 30/10 or 3.0

If fn = 30 cps, then dfr = 30/30 or 1.0

If fn = 60 cps, then dfr = 30/60 or 0.5, etc.

One way to change the fn of the valve is to vary its thickness.

CAUTION: I think the fn is proportional to the thickness raised to the 3rd power so it won't take much change in thickness to radically alter its value. IOW if you double the thickness, you will increase its resonant frequency by 8 times!

yes, that makes sense.. .. thinking about it, I realized I assumed cycles meant no flow, but pulses instead, and then just dismissed dfr>1 as an impossibility.
Cheers

p.s. I'm no good at math, but the graphs and your comments makes it almost understandable to me, and in spite of my expectations when you started this ( and the cybernetics), I have come to find it interesting. Just outta curiosity, exactly what do you mean by "a variable speed, reciprocating, positive displacement pump"? Not centrifugal, then, but a diaphagm? or did I get that one wrong as well?

The pump I envisioned was more like a miniature air compressor pump working in reverse. When the piston comes down, a check valve stops the flow of air from the environment but another port allows air to be withdrawn from my 'valve box'. When the piston comes up, it pushes some air back into the box, but the majority gets pushed back into the atmosphere since that check valve now opens. A variable speed DC motor could be coupled to this small pump.